A student connects a cell of emf `epsilon_(2)` and internal resistance `r_(2)`, with a cell of emf `epsilon_(1)` such that their combinationhas a net internal resistance less then `r_(1)`. This combination in the connected across a resistance `R`. Draw a circuit of the 'set up' and obtain an expression for the current flowing through the resistance `R`

Explanation

As the two cells are connected in parallel between the same two points `B_1` and `B_2` the potential difference V across both cells such must be same.

The potential difference between the terminals of first cell is
` V=V_B1-V_B2=epsilon_1-I_1r_1`
`therefore I_1=(epsilon_1-V)/(r_1)`
The potential difference between the terminals of `epsilon_2` is
` V=V_B1-V_B2=epsilon_2-I_2r_2`
`I_2 (epsilon_2-V)/(r_2)`
Hence, ` I= I_1+I_2=(epsilon_1-V)/(r_1)+(epsilon_2-V)/(r_2)=((epsilon_1)/(r_1)+(epsilon_2)/(r_2))-V((1)/(r_1)+(1)/(r_2))`
or ` V((r_1+r_2)/(r_1+r_2))=(epsilon_1r_2+epsilon_2r_1)/(r_1r_2)-I`
` V= (epsilon_1r_2+epsilon_2r_1)/(r_1+r_2)-I(r_1r_2)/(r_1+r_2)`.